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Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$?

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    $\begingroup$ Hint: How can one scale a matrix to make its determinant 1? Why does this work over $\mathbb{C}$ but not over $\mathbb{R}$? $\endgroup$ – user641 Mar 18 '12 at 6:52
  • $\begingroup$ I can show that first part but I don't have idea for second part $\endgroup$ – Babak Miraftab Mar 18 '12 at 10:40
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One way of proving the non-isomorphism part would be to show that ${\rm PGL}(2,{\mathbb R})$ has the Klein 4-group $C_2 \times C_2$ as subgroup, but ${\rm PSL}(2,{\mathbb R})$ does not.

The first claim is easy. $C_2 \times C_2$ is the image in ${\rm PGL}(2,{\mathbb R})$ of the dihedral group of order 8 generated by $\left(\begin{array}{rr}1&0\\0&-1\end{array}\right)$ and $\left(\begin{array}{rr}0&1\\1&0\end{array}\right)$.

It is also straightforward to show that the only element of order 2 in ${\rm SL}(2,{\mathbb R})$ is $-I_2$. So the only possibility for the inverse image in ${\rm SL}(2,{\mathbb R})$ of $C_2 \times C_2$ is the quaternion group $Q_8$. But $Q_8$ does not have a 2-dimensional real representation. That can be shown using the Frobenius-Schur indicator of the 2-dimensional complex representation, but I expect you would prefer a more elementary proof.

What is the source of this problem? Is it an exercise, and if so at what level?

Edit: In fact it is not hard to show that $Q_8$ is not a subgroup of ${\rm GL}(2,{\mathbb R})$ without using representation theory. An element of order 4 in ${\rm GL}(2,{\mathbb R})$ has minimal polynomial $x^2+1$ and is therefore conjugate to $A := \left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$. By simple linear algebra we find that the matrices conjugating $A$ to $A^{-1}$ have the form $B:=\left(\begin{array}{rr}a&b\\b&-a\end{array}\right)$, and $B^2 = -I$ gives $a^2+b^2 = -1$, which has no solution in ${\mathbb R}$.

So you now have three proofs of the non-isomorphism!

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  • $\begingroup$ I have studied linear groups.it is exercise. $\endgroup$ – Babak Miraftab Mar 18 '12 at 17:39
  • $\begingroup$ I don't want to representation theory I'm looking for group theory method $\endgroup$ – Babak Miraftab Mar 18 '12 at 18:21
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    $\begingroup$ Marc van Leeuwen's argument that ${\rm PGL}(2,{\mathbb R})$ has a subgroup of index 2 but ${\rm PSL}(2,{\mathbb R})$ does not is probably the best method. $\endgroup$ – Derek Holt Mar 18 '12 at 18:35
  • $\begingroup$ Perhaps an easier argument is that $PSL(2,\mathbb{R})$ really only has one conjugacy class of involutions: $\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}$. However, $PGL(2,\mathbb{R})$ has another: $\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$. $\endgroup$ – user641 Mar 18 '12 at 20:18
  • $\begingroup$ @steve,why has PSL(2,R) only one conjugacy class? $\endgroup$ – Babak Miraftab Mar 19 '12 at 6:51
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In the $\mathbb{C}$ case, let us consider an invertible matrix $X\in GL(2,\mathbb{C})$. Now because $\mathbb{C}$ is algebraically closed, we can choose $z\in\mathbb{C}$ such that $z^2=\det(X)$. Then of course if $Z=zI$ we have $X=ZY$, where $\det(Y)=1$; thus $Y\in SL(2,\mathbb{C})$. In particular, the inclusion $SL(2,\mathbb{C})\rightarrow GL(2,\mathbb{C})$, followed by the projection $GL(2,\mathbb{C})\rightarrow PGL(2,\mathbb{C})$, induces an isomorphism $PSL(2,\mathbb{C})\cong PGL(2,\mathbb{C})$, since $Z$ is scalar.

In the $\mathbb{R}$ case, note that the only element of order $2$ in $SL(2,\mathbb{R})$ is $\begin{pmatrix} -1 & 0\\ 0 & -1\end{pmatrix}$. Thus elements of order $2$ in $PSL(2,\mathbb{R})$ come from elements of order $4$ in $SL(2,\mathbb{R})$. Such elements satisfy $x^4-1=0$, so that their characteristic polynomial divides $x^4-1$. And since their determinant is one, we must have constant term (in their characteristic polynomial) equal to $1$. The only possibility then is $x^2+1$. Considering rational canonical form, every such matrix is conjugate (in $SL(2,\mathbb{R})$, hence in $PSL(2,\mathbb{R})$) to $A=\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}$. Thus $PSL(2,\mathbb{R})$ has one conjugacy class of involutions.

However, $B=\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$ lives in $PGL(2,\mathbb{R})$ as an element of order 2. And of course so does $A$. To see they are not conjugate, suppose there existed an invertible matrix $M$ and a scalar matrix $Z=zI$ such that $BM=AZ$. Taking determinants, we have $-1=1*z^2$, impossible over $\mathbb{R}$. Thus $PSL(2,\mathbb{R})\not\cong PGL(2,\mathbb{R})$.

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    $\begingroup$ I think the conjugacy class of elements of order 4 in ${\rm GL}(2,{\mathbb R})$ splits into two classes in ${\rm SL}(2,{\mathbb R})$ with representatives $A$ and $-A$. But that's OK, because they both map onto the same element of ${\rm PSL}(2,{\mathbb R})$. $\endgroup$ – Derek Holt Mar 19 '12 at 17:54
  • $\begingroup$ Ahh, yes, you are right. $\endgroup$ – user641 Mar 19 '12 at 20:53
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You have surjective morphisms $xL(n,K)\to PxL(n,K)$ (whose kernel consists of the multiples of the identity) for $x\in\{G,S\}$, $n\in\mathbb N$ and and $K\in\{\mathbb C,\mathbb R\}$. You also have embeddings $SL(n,K)\to GL(n,K)$. Since the kernel of the composed morphism $SL(n,K)\to GL(n,K)\to PGL(n,K)$ contains (and in fact coincides with) the kernel of the morphism $SL(n,K)\to PSL(n,K)$ (it is the (finite) set of multiples of the identity in $SL(n,K)$), one may pass to the quotient to obtain a morphism $PSL(n,K)\to PGL(n,K)$, which is injective (because of "coincides with" above). The question is whether this morphism is surjective.

The question amounts to the following: given and $g\in GL(n,K)$, does its image in $PGL(n,K)$ coincide with the image of some $g'\in SL(n,K)\subset GL(n,K)$? For that to happen, there should be a $\lambda\in K^\times$ such that $\lambda g\in SL(n,K)$, and since $\det(\lambda g)=\lambda^n\det g$, one is led to search for solutions $\lambda$ of $\lambda^n=(\det g)^{-1}$, where $(\det g)^{-1}$ could be any element of $K^\times$. It is easy to see how the solvability of this polynomial equation depends on $n$ and $K$.

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  • $\begingroup$ I don't see how you can prove the non-isomorphism part with that argument. $\endgroup$ – Derek Holt Mar 18 '12 at 11:06
  • $\begingroup$ @Marc van leeuwen what are L(n,K) and G and S? $\endgroup$ – Babak Miraftab Mar 18 '12 at 11:12
  • $\begingroup$ $G$ and $S$ are just letters, and $xL(n,K)$ denotes $GL(n,K)$ or $SL(n,K)$. $\endgroup$ – Derek Holt Mar 18 '12 at 11:45
  • $\begingroup$ @DerekHolt: You are right in asking about non-isomorphism, in the sense that the argument given will only establish that the natural morphism $PSL(n,K)\to PGL(n,K)$ it constructs is non-injective, not that no isomorphism can exist. However once you know what's absent from the image of the natural morphism, you know where to look. Since the center of $GL(2m,\mathbb R)$ contains matrices of positive determinant only, $PGL(2m,\mathbb R)$ has two connected components, but $PSL(2m,\mathbb R)$ is connected. $\endgroup$ – Marc van Leeuwen Mar 18 '12 at 12:46
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    $\begingroup$ And to dispell the spectre of a possible non-continuous isomorphism: $PGL(2m,\mathbb R)$ has an index-$2$ subgroup (isomorphic to $PSL(2m,\mathbb R)$) but $PSL(2m,\mathbb R)$ is a simple group, so it has no index-$2$ subgroups. $\endgroup$ – Marc van Leeuwen Mar 18 '12 at 12:55
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Hint: Let $K$ be any field. $PSL_{n}(K) \simeq PGL_{n}(K)$ iff $SL_{n}(K)$ intersects all cosets of $Z(GL_{n}(K))$ in $GL_{n}(K)$

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The determinant map gives rise to a split exact sequence

$$ PSL(n, F) \hookrightarrow PGL(n,F) \twoheadrightarrow F^\times / (F^\times)^n, $$

i.e. we have an isomorphism $PSL(n,F) \rtimes F^\times / (F^\times)^n \cong PGL(n,F)$.

Now, $F^\times / (F^\times)^n = 1$, if and only if any element of $F$ has a $n$-th root in it.

Since $\mathbb{R}$ is not algebraically closed, $F^\times / (F^\times)^2$ has two elements. But the complex numbers are algebraically closed.

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  • $\begingroup$ I think you wanted to answer the other question :) $\endgroup$ – t.b. Mar 19 '12 at 10:35
  • $\begingroup$ I did see the other question not before answering this one. But yes, it applies equally well to the other question, so I have copied and pasted my answer there as well. $\endgroup$ – Marc Palm Mar 19 '12 at 13:49
  • $\begingroup$ «[...] if and only if $H$ has the $n$th root in it» does not make sense. The $n$th root of what? (The next time you see that an answer answers two questions, make a link to make this explicit: otherwise there is useless duplication of content, and of fixes...) $\endgroup$ – Mariano Suárez-Álvarez Mar 20 '12 at 0:42

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